Ta có:a/b=b/c=c/a (1)
=>(a2.c)/abc=(b2.a)/abc=(c2.b)/abc(quy đồng)
=>a2c=b2a=c2.b
Ta lại có: a2c=b2a=>a2/a=b/c=>a=b/c (2)
b2a=c2.b=>b2/b=c/a=>b=c/a (3)
a2c=c2.b=>c2/c=a/b (4)
Từ (1), (2), (3),(4) => a=b=c (.)
Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a} \)<1>
=> \(\dfrac{(a^2.c)}{abc}\)=\(\dfrac{(b^2.a)}{abc} \)=\(\dfrac{(c^2.b)}{abc}\)
=> a2.c = b2.a = c2.b
Và a2.c = b2.a => \(\dfrac{a^2}{a} \)=\(\dfrac{b}{c} \)=> a= \(\dfrac{b}{c}\)<2>
b2.a = c2.b => \(\dfrac{b^2}{b}=\dfrac{c}{a}\)=> b= \(\dfrac{c}{a}\)<3>
a2.c = c2.b => \(\dfrac{c^2}{c}=\dfrac{a}{b}\)<4>
Từ <1><2><3><4> => a=b=c